二叉树四种遍历非递归

公告

Posted on 2014-11-26 20:45 Lyqatdl 阅读(156) 评论(0) 收藏举报

 1 void levelOrder(Bitree* root){
 2     queue<Node*> nodeQueue;
 3     Node* pointer=root;
 4     if(pointer){
 5         nodeQueue.push(pointer);
 6     }
 7     while(!nodeQueue.empty()){
 8         pointer=nodeQueue.front();
 9         visit(pointer);
10         nodeQueue.pop();
11         if(pointer->left)nodeQueue.push(pointer->left);
12         if(pointer->right)nodeQueue.push(pointer->right);
13     }
14 }

广度遍历，用队列存节点，访问后左右子节点入列，原节点弹出

 1 void preOrder(Bitree* root){
 2     stack<Node*> nodeStack;
 3     Node* pointer=root;
 4     while(!nodeStack.empty()||pointer){
 5         if(pointer){
 6             visit(pointer);
 7             if(pointer->right==NULL) nodeStack.push(pointer->right);
 8             pointer=pointer->left;
 9         }
10         else{
11             pointer=nodeStack.top();
12             nodeStack.pop();
13         }
14     }
15 }

先序遍历，向左扫，有右节点则右节点入栈，扫完一撇后指向栈顶，再扫一撇

 1 void inOrder(Bitree* root){
 2     stack<Node*> nodeStack;
 3     Node* pointer=root;
 4     while(!nodeStack.empty()||pointer){
 5         if(pointer){
 6             nodeStack.push(pointer);
 7             pointer=pointer->left;
 8         }
 9         else{
10             pointer=nodeStack.top();
11             visit(pointer);
12             pointer=pointer->right;
13             nodeStack.pop();
14         }
15     }
16 }

中序遍历，先一撇节点入栈，直到NULL，然互扫栈顶，指向右节点继续

 1 void posOrder(Bitree* root){
 2     stack<Node*> nodeStack;
 3     Node* pointer=root;
 4     Node* pre=root;
 5     while(pointer){
 6         for(;pointer->left!=NULL;pointer=pointer->left) nodeStack.push(pointer);
 7         while(pointer!=NULL&&(pointer->right==NULL||pointer->right==pre)){
 8             visit(pointer);
 9             pre=pointer;
10             if(nodeStack.empty()) return;
11             pointer=nodeStack.top();
12             nodeStack.pop();
13         }
14         nodeStack.push(pointer);
15         pointer=pointer->right;
16     }
17 }

后序遍历，先一撇入栈，然后如果节点右儿子扫完或没右儿子，扫节点，指向栈顶；如果有右儿子，节点入栈，指向右儿子继续。

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